Woodward — The Efficiency of Gearing under Friction. 



99 



Ratio of Energy Lost to Energy Exerted. The whole 

 energy lost during the action of one pair of teeth is £7] + U 2 . 



The total energy exerted by the driver during the same 



time is M 1 multiplied by the arc (radius measure) described 



by the driver during the action. If we let the wheels have n x 



2?r 

 and n 2 teeth, the arc required is — , and hence 



n 



U=M, 



i 

 2tt 



(7) 



The ratio of work lost to energy exerted (which ratio I will 

 call It) is 



R 



u 



But the co-efficient 



r x \r x rj \rj \n x nj l 



Substituting this, and (5), (6) and (7), we get 



, a n, of— log e (sin0.— Arcosfl.) — k("— 6,) 



Uij nj it \r J 



+ 



1 +k 2 



iog e (sin e 2 + v cos e 2 ) + v (| _ e 2 ) 



1 + £' 2 



(8) 



in which 



k= 1 



(1-?)/. * -(! + ?)/■ 



The efficiency of the combination is found by subtracting R 

 from unity. 



Efficiency = 1 — R (9). 



Formula (8) is general and exact when but one pair of 

 teeth is in action at a time. 



7. In applying (9) to an ideal case the sum of d L and 6 

 must be calculated from the equation 



